Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.9$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $6.7$ years.
$13.9$ $11.5$ $16.3$ $9.1$ $18.7$ $6.7$ $21.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $13.9$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $11.5$ years and one standard deviation above the mean is $16.3$ years. Two standard deviations below the mean is $9.1$ years and two standard deviations above the mean is $18.7$ years. Three standard deviations below the mean is $6.7$ years and three standard deviations above the mean is $21.1$ years. We are interested in the probability of a lion living less than $6.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $6.7$ years and the other half $({0.15\%})$ will live longer than $21.1$ years. The probability of a particular lion living less than $6.7$ years is ${0.15\%}$.